It takes 200 J of work to compress quasi-statically 0.40 mol of an ideal gas to one-fourth its original volume. Calculate the temperature of the gas (in K), assuming it remains constant during the compression.

Answer : The temperature of the gas is, 43.4 K

Explanation :

The expression used for work done will be,

where,

w = work done = 200 J

n = number of moles of gas  = 0.40 mol

R = gas constant = 8.314 J/mole K

T = temperature of gas  = ?

= initial volume of gas  = X

= final volume of gas  =

Now put all the given values in the above formula, we get :

Thus, the temperature of the gas is, 43.4 K

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases . this process does not change the mass number.

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as -decay. in this the mass number remains same.

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.